4 boys and 2 girls occupy seats in a row at r | vedclass

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(C) Total number of people = $4 + 2 = 6$. The total number of ways to arrange $6$ people in a row is $n(S) = 6! = 720$. To find the number of ways whe

Vedclass · Accepted Answer

$\frac{1}{3}$

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(C) Total number of people = $4 + 2 = 6$. The total number of ways to arrange $6$ people in a row is $n(S) = 6! = 720$. To find the number of ways where the $2$ girls sit side by side,we treat the $2$ girls as a single unit. Now,we have $4$ boys and $1$ unit (the $2$ girls),which makes $5$ units in total. These $5$ units can be arranged in $5!$ ways. The $2$ girls within their unit can be arranged in $2!$ ways. So,the number of favorable outcomes is $n(E) = 5! 	imes 2! = 120 	imes 2 = 240$. The probability $P$ is given by $\frac{n(E)}{n(S)} = \frac{5! 	imes 2!}{6!} = \frac{120 	imes 2}{720} = \frac{240}{720} = \frac{1}{3}$.

$A$. Probability that none of the balls is white	$I$. $\frac{1}{70}$
$B$. Probability of getting $2$ white and $1$ blue ball	$II$. $\frac{6}{35}$
$C$. Probability of getting $2$ blue and $1$ white ball	$III$. $\frac{13}{20}$
$D$. Probability of getting $1$ red,$1$ white and $1$ blue ball	$IV$. $\frac{1}{10}$

$4$ boys and $2$ girls occupy seats in a row at random. The probability that the two girls occupy seats side by side is:

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